AS June 2023 Paper 1 Q9
9 The matrix \(\mathbf{M}\) represents the transformation T and is given by
\[\mathbf{M} = \begin{bmatrix} 3p + 1 & 12 \\ p + 2 & p^2 - 3 \end{bmatrix}\](a) In the case when \(p = 0\) show that the image of the point \((4, 5)\) under T is the point \((64, -7)\) [2 marks]
(b) In the case when \(p = -2\) find the gradient of the line of invariant points under T [3 marks]
(c) Show that \(p = 3\) is the only real value of \(p\) for which \(\mathbf{M}\) is singular. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms the product \(\mathbf{M}\begin{bmatrix} 4 \\ 5 \end{bmatrix}\) or \(\mathbf{M}^{-1}\begin{bmatrix} 64 \\ -7 \end{bmatrix}\) May use the letter \(\mathbf{M}\), or \(\mathbf{M}\) in terms of \(p\), or with \(p = 0\) | M1 | 1.1a |
| Completes a reasoned argument to prove the required result. Condone no conclusion. | R1 | 2.1 |
| (2) |
Typical solution
When \(p = 0\), then
\[\begin{aligned}\mathbf{M}\begin{bmatrix} 4 \\ 5 \end{bmatrix} &= \begin{bmatrix} 1 & 12 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} \\ &= \begin{bmatrix} 64 \\ -7 \end{bmatrix}\end{aligned}\]\(\therefore\) the image of \((4, 5)\) is \((64, -7)\)
| Scheme | Marks | AO |
|---|---|---|
| Multiplies \(\mathbf{M}\) (or \(\mathbf{M}^{-1}\)) by \(\begin{bmatrix} x \\ y \end{bmatrix}\) and equates to \(\begin{bmatrix} x \\ y \end{bmatrix}\) PI Accept \(y\) replaced with \(mx\) or \(mx + c\) | M1 | 1.1a |
| Multiplying the top row of \(\mathbf{M}\) by their \(\begin{bmatrix} x \\ y \end{bmatrix}\) | M1 | 1.1a |
| Obtains correct gradient | A1 | 1.1b |
| (3) |
Typical solution
\[\begin{bmatrix} -5 & 12 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}\]\[\therefore \ -5x + 12y = x \quad \text{and} \quad 0x + y = y\]\[12y = 6x\]\[y = \frac{1}{2}x\]The gradient is \(\dfrac{1}{2}\)
| Scheme | Marks | AO |
|---|---|---|
| Uses or states \(\det\mathbf{M} = 0\) | B1 | 3.1a |
| Forms an expression for \(\det\mathbf{M}\) in \(p\) or Substitutes \(p = 3\) and evaluates \(\det\mathbf{M}\) Condone \(ad + bc\) | M1 | 1.1a |
| Obtains a correct expression for \(\det\mathbf{M}\) in terms of \(p\) | A1 | 1.1b |
| Obtains a correct simplified equation for \(\det\mathbf{M} = 0\) in terms of \(p\) | A1 | 1.1b |
| Uses a correct method to deduce that \(\det\mathbf{M} = 0\) has exactly one real root | M1 | 2.2a |
| Completes a reasoned argument that \(p = 3\) is the only real value of \(p\) for which \(\mathbf{M}\) is singular. | R1 | 2.1 |
| (6) | ||
| (11 marks) |
Typical solution
\[\det\mathbf{M} = 0\]\[\begin{aligned}\det\mathbf{M} &= (3p + 1)(p^2 - 3) - 12(p + 2) \\ &= 3p^3 + p^2 - 9p - 3 - 12p - 24 \\ &= 3p^3 + p^2 - 21p - 27\end{aligned}\]\(\therefore\) \(\mathbf{M}\) is singular when
\[3p^3 + p^2 - 21p - 27 = 0\]\[\Rightarrow p = 3 \text{ or } p = \frac{-5 \pm \mathrm{i}\sqrt{2}}{3}\]\(\therefore\) \(p = 3\) is the only real value of \(p\) for which \(\mathbf{M}\) is singular